Rationalise the denominator of each of the following.

Solution:

(i) $\frac{1}{\sqrt{7}}$

On multiplying the numerator and denominator of the given number by $\sqrt{7}$, we get:

$\frac{1}{\sqrt{7}}=\frac{1}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}}=\frac{\sqrt{7}}{7}$

(ii) $\frac{\sqrt{5}}{2 \sqrt{3}}$

On multiplying the numerator and denominator of the given number by $\sqrt{3}$, we get:

$\frac{\sqrt{5}}{2 \sqrt{3}}=\frac{\sqrt{5}}{2 \sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{15}}{6}$

(iii) $\frac{1}{2+\sqrt{3}}$

On multiplying the numerator and denominator of the given number by $\sqrt{3}$, we get:

$\frac{1}{2+\sqrt{3}}=\frac{1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}}$

$=\frac{2-\sqrt{3}}{(2)^{2}-(\sqrt{3})^{2}}$

$=\frac{2-\sqrt{3}}{4-2}$

$=\frac{2-\sqrt{3}}{1}$

$=2-\sqrt{3}$

(iv) $\frac{1}{\sqrt{5}-2}$

On multiplying the numerator and denominator of the given number by $\sqrt{5}+2$, we get:

$\frac{1}{\sqrt{5}-2}=\frac{1}{\sqrt{5}-2} \times \frac{\sqrt{5}+2}{\sqrt{5}+2}$

$=\frac{\sqrt{5}+2}{(\sqrt{5})^{2}-(2)^{2}}$

$=\frac{\sqrt{5}+2}{5-4}$

$=\frac{\sqrt{5}+2}{1}$

$=\sqrt{5}+2$

(v) $\frac{1}{5+3 \sqrt{2}}$

On multiplying the numerator and denominator of the given number by $5-3 \sqrt{2}$, we get:

$\frac{1}{5+3 \sqrt{2}}=\frac{1}{5+3 \sqrt{2}} \times \frac{5-3 \sqrt{2}}{5-3 \sqrt{2}}$

$=\frac{5-3 \sqrt{2}}{(5)^{2}-(3 \sqrt{2})^{2}}$

$=\frac{5-3 \sqrt{2}}{25-18}$

$=\frac{5-3 \sqrt{2}}{7}$

(vi) $\frac{1}{\sqrt{7}-\sqrt{6}}$

Multiplying the numerator and denominator by $\sqrt{7}+\sqrt{6}$, we get

$\frac{1}{\sqrt{7}-\sqrt{6}}=\frac{1}{\sqrt{7}-\sqrt{6}} \times \frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}$

$=\frac{\sqrt{7}+\sqrt{6}}{(\sqrt{7})^{2}-(\sqrt{6})^{2}}$

$=\frac{\sqrt{7}+\sqrt{6}}{7-6}$

$=\sqrt{7}+\sqrt{6}$

(vii) $\frac{4}{\sqrt{11}-\sqrt{7}}$

Multiplying the numerator and denominator by $\sqrt{11}+\sqrt{7}$, we get

$\frac{4}{\sqrt{11}-\sqrt{7}}=\frac{4}{\sqrt{11}-\sqrt{7}} \times \frac{\sqrt{11}+\sqrt{7}}{\sqrt{11}+\sqrt{7}}$

$=\frac{4(\sqrt{11}+\sqrt{7})}{(\sqrt{11})^{2}-(\sqrt{7})^{2}}$

$=\frac{4(\sqrt{11}+\sqrt{7})}{11-7}$

$=\sqrt{11}+\sqrt{7}$

(viii) $\frac{1+\sqrt{2}}{2-\sqrt{2}}$

Multiplying the numerator and denominator by $2+\sqrt{2}$, we get

$\frac{1+\sqrt{2}}{2-\sqrt{2}}=\frac{1+\sqrt{2}}{2-\sqrt{2}} \times \frac{2+\sqrt{2}}{2+\sqrt{2}}$

$=\frac{2+\sqrt{2}+2 \sqrt{2}+2}{(2)^{2}-(\sqrt{2})^{2}}$

$=\frac{4+3 \sqrt{2}}{4-2}$

$=\frac{4+3 \sqrt{2}}{2}$

(ix) $\frac{3-2 \sqrt{2}}{3+2 \sqrt{2}}$

Multiplying the numerator and denominator by $3-2 \sqrt{2}$, we get

$\frac{3-2 \sqrt{2}}{3+2 \sqrt{2}}=\frac{3-2 \sqrt{2}}{3+2 \sqrt{2}} \times \frac{3-2 \sqrt{2}}{3-2 \sqrt{2}}$

$=\frac{(3-2 \sqrt{2})^{2}}{(3)^{2}-(2 \sqrt{2})^{2}}$

$=\frac{9+8-12 \sqrt{2}}{9-8} \quad\left[(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$

$=17-12 \sqrt{2}$